Limit of Expectation values involving exponential i.i.d random variables

Question

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of independent and identically distributed (i.i.d) random variables with $\mathbb{E}[X_1] = 1$ and $\mathbb{V}[X_1] = 1$. Show that $$\lim_{n \rightarrow \infty} \mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2n} \bigl(\sum^{n}_{k=1} (X_k-1) \bigr)^2 \Bigr) \Bigr] = \frac{1}{\sqrt{2}}.$$ I am stuck on this problem not knowing how to approach it. Is there a hidden application of some theorem or is the proof elementary?
I would appreciate any tips on how to tackle this limit.

Edit: Based on Oscar's answer I wish to write down my attempt.
Let $\mu = \mathbb{E}[X_1] = 1$ and $\sigma^2 = \mathbb{V}[X_1] = 1>0$, then we may examine the exponent and observe that $$-\frac{1}{2n} \Bigl(\sum^{n}_{k=1} (X_k-1) \Bigr)^2 = -\frac{1}{2} \Bigl( \frac{\sum^{n}_{k=1}X_k-n}{\sqrt{n}} \Bigr)^2 = -\frac{1}{2} \Bigl( \frac{\sum^{n}_{k=1}X_k-n \mu}{\sqrt{n} \sigma} \Bigr)^2 =: -\frac{1}{2} Y_n^2$$ where we define the standardized variable $Y_n = \frac{\sum^{n}_{k=1}X_k-n \mu}{\sqrt{n} \sigma}$ for $n \in \mathbb{N}$.
The Central Limit Theorem is applicable and yields $Y_n \rightarrow Y \sim \mathcal{N}(0,1)$ in distribution, i.e. $F_{Y_n}(t) \rightarrow F_Y(t) \, \, (n \rightarrow \infty)$ for all $t \in \mathbb{R}$.
Now, we calculate the expectation value: $$\mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2} Y^2 \Bigr) \Bigr] = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp \Bigl(-\frac{1}{2} y^2 \Bigr) p(y) dy = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp \Bigl(-\frac{1}{2} y^2 \Bigr) \exp \Bigl(-\frac{1}{2} y^2 \Bigr) dy = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp \Bigl(-y^2 \Bigr) dy = \frac{1}{\sqrt{2\pi}} \sqrt{\pi} = \frac{1}{\sqrt{2}}$$

In order to finish the proof we need the statement: $$\lim_{n \rightarrow \infty} \mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2} Y^2_n \Bigr) \Bigr] = \mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2} Y^2 \Bigr) \Bigr].$$

Questions: Does this hold true?
And in general: If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a Borel-measurable function, when does $$\lim_{n \rightarrow \infty} \mathbb{E} \Bigl[ f(Y_n) \Bigr] = \mathbb{E} \Bigl[ f(Y) \Bigr]$$ hold. What conditions must $f$ satisfy?

Answer 1

Note that $$-\frac{1}{2n} \Big( \sum_{k=1}^n (X_k - 1) \Big)^2 = -\frac{1}{2} \Big( \sum_{k=1}^n \frac{(X_k - 1)}{\sqrt{n}} \Big)^2 = -\frac{1}{2} \Big( \sqrt{n} \sum_{k=1}^n \frac{(X_k - 1)}{n} \Big)^2,$$ and by the Central Limit theorem you have that $$\sqrt{n} \sum_{k=1}^n \frac{(X_k - 1)}{n} \overset{d}{\to} \mathcal{N}(0,1).$$ Also, recall moment generating functions.

--- Edit ---

To justify the final step you can use the following result:

Let $X_1,X_2,\dots$ be a sequence of random variables and suppose that $X_n \to X$ in distribution as $n\to \infty$. If $h$ is a real valued or complex valued, bounded, continuous function, then $$\mathbb{E}[h(X_n)] \to \mathbb{E}[h(X)] \quad \text{as} \ n \to \infty.$$

You can find a proof of this result in Allan Gut's Probability: A Graduate Course, Theorem 5.7. In your case, $h = e^{-x^2/2}$, which is of course bounded and continuous.

Answer 2

Let $S_{n}=\sum_{k=1}^{n}(X_{k}-1)$.

Then $S_{n}$ is the sum of iid mean $0$ random variables with variance $1$.

The Central Limit Theorem implies $\frac{S_{n}}{\sqrt{n}}\xrightarrow{d}N(0,1)$.

One very useful, and often the standard characterization of convergence in distribution is the following functional analytic one:-

We say $Z_{n}\xrightarrow{d}Z$ if for all $f\in C_{b}(\Bbb{R})$, i.e. the space of bounded continuous functions of $\Bbb{R}$, we have $$E(f(Z_{n}))\xrightarrow{n\to\infty}E(f(Z))$$

Now, define $\displaystyle f(x)=e^{-\frac{x^{2}}{2}}$ for $x\in\Bbb{R}$ . It is obvious to see that $f$ is a continuous and bounded function.

Now, we set $Z_{n}=\frac{S_{n}}{\sqrt{n}}$ and $Z=N(0,1)$.

So, the above characterization says precisely that:-

$$ \mathbb{E} \Bigl[ \exp \Bigl(-\frac{1}{2n} \bigl(\sum^{n}_{k=1} (X_k-1) \bigr)^2 \Bigr) \Bigr]=\mathbb{E}\bigg(\frac{-S_{n}^{2}}{2n}\bigg)=E(f(Z_{n}))\to E(f(Z))$$

So all you need to do is compute that

$$E(f(Z))=\int_{-\infty}^{\infty}e^{-\frac{z^{2}}{2}}\cdot\dfrac{e^{-\frac{z^{2}}{2}}}{\sqrt{2\pi}}\,dz =\frac{1}{\sqrt{2}}$$